Do you agree that, for $\displaystyle x=\emptyset,$ your formula is true?

Now assume that it is true for a $\displaystyle x\in\omega,$ we have to show that it's also true for its successor, $\displaystyle x\cup\{x\}.$

Let $\displaystyle y$ be an element of $\displaystyle x\cup\{x\},$ i.e. $\displaystyle y\in x\ \text{or}\ y\in \{x\}.$ In the first case, the induction hypothesis states that $\displaystyle y\in\omega.$ In the second case, $\displaystyle y=x$ and then $\displaystyle y\in\omega .$

So, using the mathematical induction principle, we proved that $\displaystyle \omega$ is a transtive set.

Feb 22nd 2009, 07:10 AM

ThePerfectHacker

Here is a non-induction way to prove this. I know you asked for induction but I bring this approach up because it might help you when you do more stuff on ordinals. If $\displaystyle X$ is a set of ordinals then $\displaystyle \cup X$ is a set of ordinals. Since $\displaystyle \omega$ is the union of all natural numbers and since each natural number is an ordinal it follows that $\displaystyle \omega$ is an ordinal.

Feb 23rd 2009, 06:29 AM

georgel

Thanks to both of you, I will probably have another question or two in the next couple of days.. :)